Question: Is ${78990}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {78990}= &&{7}\cdot10000+ \\&&{8}\cdot1000+ \\&&{9}\cdot100+ \\&&{9}\cdot10+ \\&&{0}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {78990}= &&{7}(9999+1)+ \\&&{8}(999+1)+ \\&&{9}(99+1)+ \\&&{9}(9+1)+ \\&&{0} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {78990}= &&\gray{7\cdot9999}+ \\&&\gray{8\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {7}+{8}+{9}+{9}+{0} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first four terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${78990}$ is divisible by $9$ if ${ 7}+{8}+{9}+{9}+{0}$ is divisible by $9$ Add the digits of ${78990}$ $ {7}+{8}+{9}+{9}+{0} = {33} $ If ${33}$ is divisible by $9$ , then ${78990}$ must also be divisible by $9$ ${33}$ is not divisible by $9$, therefore ${78990}$ must not be divisible by $9$.